Mole concept There are 6,02 x 10 23 atoms in 12 g of carbon-12. - - PowerPoint PPT Presentation

Mole concept

There are 6,02 x 1023 atoms in 12 g of carbon-12. Avogadro’s constant (NA)

1 Mole is the amount of substance containing as many particles (atoms, ions, molecules) as there are atoms found in 12 g carbon-12.

One mole of atoms is 6,02 x 1023 atoms One mole of ions is 6,02 x 1023 ions One mole of molecules is 6,02 x 1023 molecules

How many atoms are found in 1 mole of argon atoms?

6,02 x 1023 atoms

How many molecules are found 1 mole of HCℓ molecules?

6,02 x 1023 molecules

How many atoms are found in 1 mole of HCℓ molecules?

2 x 6,02 x 1023 atoms

How many hydrogen atoms are found in 1 mole of HCℓ molecules?

6,02 x 1023 H atoms

How many chlorine atoms are found in 1 mole of HCℓ molecules?

6,02 x 1023 Cℓ atoms

How many chlorine molecules are found in 1 mole of Cℓ2 molecules?

6,02 x 1023 Cℓ2 molecules

How many chlorine atoms are found in 1 mole of Cℓ2 molecules?

2 x 6,02 x 1023 Cℓ atoms

How many water molecules are found in 1 mole of H2O molecules?

6,02 x 1023 H2O molecules

How many atoms are found in 1 mole of H2O molecules?

3 x 6,02 x 1023 atoms

How many H atoms are found in 1 mole of H2O molecules?

2 x 6,02 x 1023 H atoms

How many O atoms are found in 1 mole of H2O molecules?

6,02 x 1023 O atoms

How many moles of atoms 3,01 x 1023 atoms?

𝑜 = 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑞𝑏𝑠𝑢𝑗𝑑𝑚𝑓𝑡 𝑂

𝐵

= 3,01𝑦1023 6,02𝑦1023 = 0,5 𝑛𝑝𝑚

How many moles of H2O molecules is found in 36 g of water?

𝑜 = 𝑛 𝑁

= 36 18

= 2 𝑛𝑝𝑚

M(H2O) = 18 g.mol-1

Class work:

Exercise 1

• Pg. 141,
• Nos. 4, 8, 9, 17, 20, 21

• 4. Bereken die getal chhlooratome in 8 mol chloor (Cl2)

n =

𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑛𝑝𝑚𝑓𝑑𝑣𝑚𝑓𝑡 𝑂𝐵 number of molecules = 8 x 6,02 x 1023 Cℓ₂ molecules

= 4,82 x 1024 Cℓ₂ molecules

number of atoms = 2 x 4,82 x 1024 Cℓ atoms

= 9,63 x 1024 Cℓ atoms

• 4. Calculate the number of chlorine atoms in 8 mol of chlorine (Cℓ₂)

• 8. Calculate the amount of mole substance represented by each of the following given masses.

8.1 12 g of Mg atoms

n =

𝑛 𝑁

=

12 24

= 0,5 mol

• 8. Calculate the amount of mole substance represented by each of the following given masses.

8.2 24 g of ammonia (NH3)

n =

𝑛 𝑁

=

24 17

= 1,41 mol

• 8. Calculate the amount of mole substance represented by each of the following given masses.

8.3 6,4 g of sulphur dioxide

n =

𝑛 𝑁

=

6,4 64

= 0,1 mol

• 8. Calculate the amount of mole substance represented by each of the following given masses.

8.4 1,5 g of water

n =

𝑛 𝑁

=

1,5 18

= 0,083 mol

• 9. Calculate the mass (in gram) of each of the following amounts of substance:

9.1 0,6 mol of carbon dioxide

n =

𝑛 𝑁

0,6 =

𝑛 44

m = 26,4 g

• 9. Calculate the mass (in gram) of each of the following amounts of substance:

9.2 10 mol of HCℓ

n =

𝑛 𝑁

10 =

𝑛 36,5

m = 365 g

• 9. Calculate the mass (in gram) of each of the following amounts of substance:

9.3 0,25 mol of CaCO3

n =

𝑛 𝑁

0,25 =

𝑛 100

m = 25 g

• 9. Calculate the mass (in gram) of each of the following amounts of substance:

9.4 1 mol of NaOH

n =

𝑛 𝑁

1 =

𝑛 40

m = 40 g

• 17. Calculate the number of C atoms and H atoms in 74 g of C3O2H6.

n =

𝑛 𝑁

=

74 74

= 1 mol

number of C atoms = 3 x 1 x 6,02 x 1023 = 1,81 x 1024 atoms number of H atoms = 6 x 1 x 6,02 x 1023 = 3,61 x 1025 atoms

• 20. Calculate the number of magnesium ions and nitrate ions in 118,4 g of Mg(NO3)2.

n =

𝑛 𝑁

=

118,4 148

= 0,8 mol

number of Mg2+ ions = 1 x 0,8 x 6,02 x 1023 = 4,82 x 1023 ions number of NO3

• ions = 2 x 0,8 x 6,02 x 1023

= 9,6 x 1023 ions

• 21. Calculate the mass of 6,02 x 1023 I2 molecules.

n =

𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑛𝑝𝑚𝑓𝑑𝑣𝑚𝑓𝑡 𝑂𝐵

=

6,02×1023 6,02×1023

= 1 mol

n =

𝑛 𝑁

1 =

𝑛 254

m = 254 g

Molar gas volume One mole of any gas at standard conditions (STP) has a volume (Vm)

• f 22,4 dm3

Molar gas volume

n =

𝑊 𝑊

𝑛

Example

n =

𝑊 𝑊

𝑛

Calculate the number of mole of carbon dioxide in a 3 dm3 container at STP .

=

3 22,4

= 0,13 mol

Molar concentration When 6,02 x 1023 molecules are contained in 1 dm3, the concentration

• f the gas is 1 mol.dm-3

Molar concentration

1 dm3 = 1 dm x 1 dm x 1 dm

= 10 cm x 10 cm x 10 cm = 1 000 cm3

Molar concentration

1 mol.dm-3 2 mol.dm-3 3 mol.dm-3

Molar concentration

Therefore:

c =

𝑜 𝑊

Standard solution A solution of which the concentration is exactly known (and remains constant for some time).

Example

Mr(NaOH) = 40

Calculate the concentration of a sodium hydroxide solution if 20 g of sodium hydroxide is dissolved in 500 cm3 of water.

=

20 40 = 0,5 mol

n =

𝑛 𝑁

=

0,5 0,5 = 1 mol.dm-3

c =

𝑜 𝑊

Example – Exercise 16, pg. 146, no. 1

Calculate the concentration of a solution with a volume of 250 cm3 if it contains 73 g of HCℓ. n =

𝑛 𝑁

=

73 36,5

= 2 mol c =

𝑜 𝑊

=

2 0,25

= 8 mol.dm-3 Or c =

𝑛 𝑁𝑊 = 73 (36,5)(0,25) = 8 mol.dm-3

Percentage composition

The mass of each element in a compound is expressed as a percentage of the total mass

• f the compound.

Example (Pg. 148)

Determine the percentage composition of NaHCO3 Mr(NaHCO3) = 84 Mr(Na) = 23 Mr(H) = 1 Mr(C) = 12 Mr(O) = 16 Percentage Na =

23 84 x 100 = 27,38%

Percentage H =

1 84 x 100 = 1,19%

Percentage C =

12 84 x 100 = 14,29%

Percentage O =

16𝑌3 84 x 100 = 57,14%

Example (Pg. 149)

Determine the percentage of iron in Fe2O3 M(Fe2O3) = 160 g.mol-1 M(Fe) = 56 g.mol-1

Percentage Fe =

2𝑌56 160 x 100 = 70%

Example (Pg. 149)

Determine the percentage water of crystallisation in CuSO4.5H2O.

Mr(CuSO4.5H2O) = 249,5 Mr(H2O) = 18

Percentage H2O =

5𝑌18 249,5 x 100 = 36,07%

Power Gym: Exercise 2

• Pg. 147,
• Nos. 1.1, 1.3, 2, 8, 10, 13, 15

Empirical formula: Indicates the simplest ratio in which the elements combine to form the compound.

Example: Exercise 18, no. 1.1 Determine the empirical formula of the compound with the following percentage composition per mass: 63,5% Fe; 36,5% S

Suppose in 100 g of the compound: Fe: n =

𝑛 𝑁 = 63,5 56 = 1,13 mol

S: n = 𝑛

𝑁 = 36,5 32 = 1,14 mol

Fe : S 1,13 : 1,14 1 : 1 Thus the empirical formula is FeS.

Molecular formula: Indicates the number of atoms of each element in a compound and may be a multiple of the empirical formula.

Example: Exercise 19, no. 1 A compound has an empirical formula of CH2Cℓ and the relative molecular mass

• f the compound is 99. Determine the molecular formula.

M(CH2Cℓ) = 49,5 g.mol-1 M(actual formula) = 99 g.mol-1

𝑁(𝑏𝑑𝑢𝑣𝑏𝑚 𝑔𝑝𝑠𝑛𝑣𝑚𝑏) 𝑁(𝑓𝑛𝑞𝑗𝑠𝑗𝑑𝑏𝑚 𝑔𝑝𝑠𝑛𝑣𝑚𝑏) = 99 49,5 = 2

Thus the actual formula: C2H4Cℓ2

Homework: Exercise 18

• Pg. 154,
• Nos. 1.2, 1.3, 2

Exercise 19

• Pg. 156,
• No. 2

Balanced equations 4 Na + O2 → 2 Na2O

The number of moles of reactants and products:

Na Na Na Na O O Na Na Na Na O O 4 mol Na atoms + 1 mol O2 molecules → 2 mol Na2O bound in a crystal lattice

Limiting (limited) reagents: A reaction continues until one of the reagents is consumed. This reagent is the limited reagent and all other reagents are in excess.

Example: Use the unbalanced chemical equation to answer the questions following: SO2(g) + O2(g) → SO3(g) 3.1 Balance the equation.

2SO2(g) + O2(g) → 2SO3(g)

Example: Use the unbalanced chemical equation to answer the questions following: 2SO2(g) + O2(g) → 2SO3(g) 3.2 How many moles of SO3 will be formed when 3 mol of SO2 reacts with sufficient O2?

Ratio according to balanced equation:

2 mol of SO2 forms 2 mol of SO3 2 : 2 1 : 1 Thus 3 mol of SO2 forms 3 mol of SO3

Example: Use the unbalanced chemical equation to answer the questions following: 2SO2(g) + O2(g) → 2SO3(g) 3.3 If 6 mol of O2 is available, how much SO2 is needed to form 4 mol of SO3?

Ratio according to balanced equation:

2 mol SO2 : 1 mol O2 : 2 mol SO3 4 mol : 2 mol : 4 mol Thus 4 mol SO2 is needed

Example: Use the unbalanced chemical equation to answer the questions following: 2SO2(g) + O2(g) → 2SO3(g) 3.4 How many moles of O2 is in excess in 3.3?

Ratio according to balanced equation:

4 mol : 2 mol : 4 mol Thus (6 - 2 mol) = 4 mol O2 is in excess

Example: Use the unbalanced chemical equation to answer the questions following: 2SO2(g) + O2(g) → 2SO3(g) 3.5 If 160 g of SO2 reacts with sufficient O2, calculate the mass of SO3 formed.

M(SO2) = 64 g.mol-1

n =

𝑛 𝑁 = 160 64

= 2,5 mol

Example: Use the unbalanced chemical equation to answer the questions following: 2SO2(g) + O2(g) → 2SO3(g) 3.5 If 160 g of SO2 reacts with sufficient O2, calculate the mass of SO3 formed.

M(SO3) = 80 g.mol-1

n =

𝑛 𝑁

2,5 =

𝑛 80

→ m = 200 g SO3

From the balanced equation 2,5 mol of SO2 will form 2,5 mol of SO3

Example: Use the unbalanced chemical equation to answer the questions following: 2SO2(g) + O2(g) → 2SO3(g) 3.6 How many molecules of SO3 are formed in 3.5?

1 mol of molecules is 6,02 x 1023 molecules 2,5 mol of molecules is n x NA = 2,5 x 6,02 x 1023 molecules = 1,505 x 1024 SO3 molecules

Example: Use the unbalanced chemical equation to answer the questions following: 2SO2(g) + O2(g) → 2SO3(g) 3.7 Is it possible to produce 80 g of SO3 if only 8 g of O2 is available? Explain.

M(SO3) = 80 g.mol-1

n =

𝑛 𝑁 = 80 80 = 1 mol

Example: Use the unbalanced chemical equation to answer the questions following: 2SO2(g) + O2(g) → 2SO3(g) 3.7 Is it possible to produce 80 g of SO3 if only 8 g of O2 is available? Explain.

M(O2) = 32 g.mol-1

n =

𝑛 𝑁

0,5 =

𝑛 32

→ m = 16 g

From the balanced equation 1 mol of O2 is needed to form 2 mol of SO3 → 0,5 mol of O2 is needed to form 1 mol of SO3

→ 16 g of oxygen is needed, therefore 8 g will not be sufficient.

In 3.7 oxygen will be the limited reagent and SO2 will be in excess if there is more than 1 mol of SO2 available.

Percentage yield

Percentage yield =

𝑏𝑑𝑢𝑣𝑏𝑚 𝑧𝑗𝑓𝑚𝑒 𝑢ℎ𝑓𝑝𝑠𝑓𝑢𝑗𝑑𝑏𝑚 𝑧𝑗𝑓𝑚𝑒

Example, pg. 167

When 12,8 g of Cu is allowed to burn (combust) in oxygen, 15,2 g of copper(II) oxide is produced. Determine the percentage yield.

2Cu(s) + O2(g) → 2CuO(s) M(CuO) = 79,5 g.mol-1 M(Cu) = 63,5 g.mol-1 Cu: n =

𝑛 𝑁 = 12,8 63,5 = 0,2 mol Cu

From the balanced equation: 2 mol of Cu yields 2 mol of CuO 0,2 mol of Cu yields 0,2 mol of CuO

Example, pg. 167

When 12,8 g of Cu is allowed to burn (combust) in oxygen, 15,2 g of copper(II) oxide is produced. Determine the percentage yield.

2Cu(s) + O2(g) → 2CuO(s) n =

𝑛 𝑁

0,2 =

𝑛 79,5

m = 16,03 g (theoretical yield) Percentage yield =

𝑏𝑑𝑢𝑣𝑏𝑚 𝑧𝑗𝑓𝑚𝑒 𝑢ℎ𝑓𝑝𝑠𝑓𝑢𝑗𝑑𝑏𝑚 𝑧𝑗𝑓𝑚𝑒 x 100

=

15,2 16,03 x 100

= 94,82 %

Applications of exothermic reactions producing large quantities of gas:

Airbags: 2NaN3(s) → 2Na(s) + 3N2(g)

Only 130 g of NaN3 is needed to produce 67 dm3

• f N2 gas in 0,03 s.

Applications of exothermic reactions producing large quantities of gas:

Octane combustion in motorcar engine: 2C8H18 + 25O2 → 16CO2 + 18H2O + energy The gas moves the pistons in the engine.

Applications of exothermic reactions producing large quantities of gas:

Termionic decomposition of ammonium nitrate: 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)

Fertiliser production and explosive in mining industry.

Homework:

Work through the following questions and answers. Always FIRST try to do the question YOURSELF.

• 1. Balance the following equation:

K + O2 → K2O 4 2

Now use the balanced equation to complete the following table with each row in the table according to the same ratio as the balanced equation:

4K O2 2K2O 0,5 1,6 0,3 0,25 10 2 1 0,4 0,8 0,6 0,15 1 0,5 2,5 5

• 2. Write down the balanced equation for nitrogen reacting

with hydrogen to form ammonia:

N2 + H2 → NH3 3 2

Now use the balanced equation to complete the following table with each row in the table according to the same ratio as the balanced equation:

N2 3H2 2NH3 1,5 0,4 2 0,9 10 4,5 3 0,2 0,6 1 3 0,3 0,6 30 20

• 3. Consider the following reaction:

NH3 + O2 → NO + H2O

3.1 Balance the reaction equation.

NH3 + O2 → NO + H2O 4 6 5 4

• 3. Consider the following reaction:

4NH3 + 5O2 → 4NO + 6H2O

3.2 How many moles of oxygen is needed to react with 1 mole of ammonia? 4 mol ammonia reacts with 5 mol oxygen 1 mol ammonia reacts with (

1 4 × 5 ) = 1,25 mol oxygen

• 3. Consider the following reaction:

4NH3 + 5O2 → 4NO + 6H2O

3.3 What mass of oxygen reacted with 1 mole of ammonia according to question 3.2? M(O2) = 32 g.mol-1

n =

𝑛 𝑁

1,25 =

𝑛 32

m = 40 g

4. Ammonia is commercially prepared as follows in the Haber process:

N2 + 3H2 → 2NH3

4.1 How many moles of NH3 is formed if one mole of N2 reacts with sufficient hydrogen?

2 mol NH3

• 4. Ammonia is commercially prepared as follows in the Haber process:

N2 + 3H2 → 2NH3

4.2 How many moles of ammonia is formed when 1,806 x 1024 molecules of hydrogen reacts with sufficient nitrogen?

n =

𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑞𝑏𝑠𝑢𝑗𝑑𝑚𝑓𝑡 𝑂𝐵

=

1,806×1024 6,02×1023 = 3 mol H2

3 mol of H2 reacts with nitrogen to form 2 mol of NH3

• 4. Ammonia is commercially prepared as follows in the Haber process:

N2 + 3H2 → 2NH3

4.3 How many moles of ammonia is formed if 0,2 mol of N2 combines with 0,6 mol of H2?

1 mol of N2 reacts with 3 mol H2 to form 2 mol of NH3 0,2 mol of N2 reacts with 0,6 mol H2 to form 0,4 mol of NH3

N2 + 3H2 → 2NH3

4.4 How many moles of ammoniais formed from 6 moles of H2?

3 mol of H2 reacts to form 2 mol of NH3 6 mol of H2 reacts to form 4 mol of NH3

N2 + 3H2 → 2NH3

4.5 Calculate the volume of gas which would form from 6 mol of H2 at STP?

4 mol of NH3 forms

n =

𝑊 𝑊

𝑛

4 =

𝑊 22,4

V = 89,6 dm3

N2 + 3H2 → 2NH3

4.6 What mass of nitrogen is needed to form 8,5 g of ammonia?

M(NH3) = 17 g.mol-1

n =

𝑛 𝑁 = 8,5 17 = 0,5 mol

2 mol of ammonia is formed from 1 mol of nitrogen 0,5 mol of ammonia is formed from 0,25 mol of nitrogen

n =

𝑛 𝑁

0,25 =

𝑛 28 → m = 7 g

M(N2) = 28 g.mol-1

N2 + 3H2 → 2NH3

4.7 What is the volume of 8,5 g of ammonia at STP?

n =

𝑊 𝑊

𝑛

0,5 =

𝑊 22,4

V = 11,2 dm3

5. Methanol (CH3OH) can be burnt in oxygen to provide energy or it can decompose into hydrogen gas which can be used as fuel, according to the following reaction:

CH3OH(ℓ) → 2H2(g) + CO(g)

5.1 If 125 g of methanol decomposes, what is the theoretical yield of the hydrogen gas?

n =

𝑛 𝑁 =

125 32

= 3,91 mol

M(CH3OH) = 32 g.mol-1

nmethanol : nH₂ 1 : 2 3,91 : 7,81

nH₂ =

𝑛 𝑁

7,81 =

𝑛 2

m = 15,63 g

5. Methanol (CH3OH) can be burnt in oxygen to provide energy or it can decompose into hydrogen gas which can be used as fuel, according to the following reaction:

CH3OH(ℓ) → 2H2(g) + CO(g)

5.2 If only 13,6 g of hydrogen gas is obtained, what was the percentage yield?

% yield =

13,6 15,63 x 100

= 87 %

• 6. Consider the following balanced equation:

2Aℓ(s) + 3 Cℓ2(g) → 2 AℓCℓ3(s) A mixture of 1,5 moles of Aℓ and 3 moles of Cℓ2 are allowed to react together. 6.1 Which substance is the limiting reactant?

2 mol of Aℓ reacts with 3 mol of chlorine gas 1,5 mol of Aℓ reacts with 2,25 mol of chlorine gas OR Aℓ : Cℓ2

2 : 3 1,5 : 2,25

3 mol of chlorine gas is available and it is therefore in excess. Therefore the aluminium is the limiting reactant.

• 6. Consider the following balanced equation:

2Aℓ(s) + 3 Cℓ2(g) → 2 AℓCℓ3(s) A mixture of 1,5 moles of Aℓ and 3 moles of Cℓ2 are allowed to react together. 6.2 How many moles of AℓCℓ3 is formed?

2 mol of Aℓ forms 2 mol of AℓCℓ3 1,5 mol of Aℓ forms 1,5 mol of AℓCℓ3 OR Aℓ : Cℓ2 : AℓCℓ3 2 : 3 : 2 1,5 : 2,25 : 1,5 thus 1,5 mol of AℓCℓ3 can be produced

• 6. Consider the following balanced equation:

2Aℓ(s) + 3 Cℓ2(g) → 2 AℓCℓ3(s) A mixture of 1,5 moles of Aℓ and 3 moles of Cℓ2 are allowed to react together. 6.3 How many mole of the reactant that was in excess, was eventually left over?

There is 3 mol but only 2,25 mol of Cℓ2 is used, thus 3 mol – 2,25 mol = 0,75 mol Cℓ2

• 7. Gerda heats 50 g of CaCO3 which reacts as follows:

CaCO3 → CaO + CO2 7.1 Determine the mass of gas she captured if the CaCO3 was 100% pure.

n =

𝑛 𝑁 =

50 100 = 0,5 mol

M(CH3OH) = 100 g.mol-1 1 mol of CaCO3 forms 1 mol of CO2 0,5 mol of CaCO3 forms 0,5 mol of CO2

n =

𝑛 𝑁

0,5 =

𝑛 44

m = 22 g

• 7. Gerda heats 50 g of CaCO3 which reacts as follows:

CaCO3 → CaO + CO2 7.2 Gerda only obtains 18,2 g of CO2 gas. Calculate the percentage purity of the CaCO3.

Remember that purity will determine the yield.

% yield =

𝑏𝑑𝑢𝑣𝑏𝑚 𝑧𝑗𝑓𝑚𝑒 𝑢ℎ𝑓𝑝𝑠𝑓𝑢𝑗𝑑𝑏𝑚 𝑧𝑗𝑓𝑚𝑒

=

18,2 22 x 100

= 82,7 %

• 7. Gerda heats 50 g of CaCO3 which reacts as follows:

CaCO3 → CaO + CO2 7.3 Calculate the mass of gas Gerda can capture if the CaCO3 was only 70% pure.

Using the answer in 7.1: mass of CO2 = 70% x 22 g = 15,4 g OR (if 7.1 was not asked)

• 7. Gerda heats 50 g of CaCO3 which reacts as follows:

CaCO3 → CaO + CO2 7.3 Calculate the mass of gas Gerda can capture if the CaCO3 was only 70% pure.

70% of 50 g is 35 g n =

𝑛 𝑁 = 35 100 = 0,35 mol

n =

𝑛 𝑁

0,35 =

𝑛 44

➔ m = 15,4 g

1 mol of CaCO3 forms 1 mol of CO2 0,35 mol of CaCO3 forms 0,35 mol of CO2

8. A learner adds 10 cm3 of sulphuric acid with a concentration of 10 mol.dm-3 to water to make a solution of 250 cm3. Calculate the concentration of the diluted acid.

c₁ =

𝑜 𝑊₁

n = 0,01 x 10 n = 0,1 mol sulphuric acid c₂ =

𝑜 𝑊₂

c₂ =

0,1 0,25

= 0,4 mol.dm-3

8. A learner adds 10 cm3 of sulphuric acid with a concentration of 10 mol.dm-3 to water to make a solution of 250 cm3. Calculate the concentration of the diluted acid. Alternative method:

𝑜𝑗 = 𝑜𝑔 c1V1 = c2V2 10 x 0,01 = c2 x 0,25 c2 = 0,4 mol.dm-3

• 5. Consider the reaction:

Aℓ2O3 + 3H2 → 2Aℓ + 3H2O

Determine the mass of Aℓ2O3 that has to react with an excess of hydrogen to prepare 135 g of Aℓ.

M(Aℓ) = 27 g.mol-1

n =

𝑛 𝑁 = 135 27 = 5 mol

2 mol of aluminium is formed from 1 mol of Aℓ2O3 5 mol of aluminium is formed from 2,5 mol of Aℓ2O3

n =

𝑛 𝑁

2,5 =

𝑛 102

→ m = 255 g

M(Aℓ2O3) = 102 g.mol-1

9.1 Aℓ : Cℓ2 2 : 3 1,5 : 2,25 If 1,5 mol of Aℓ is used, only 2,25 mol of Cℓ2 would be used. Aℓ is the limited reactant. 9.2 Aℓ : Cℓ2 : AℓCℓ3 2 : 3 : 2 1,5 : 2,25 : 1,5 thus 1,5 mol of AℓCℓ3 can be produced 9.3 We have 3 mol, but only 2,25 mol of Cℓ2 is used, thus 3 – 2,25 = 0,75 mol is left

10.1 NaOH : CO2 : Na2CO3 : H2O 2 : 1 : 1 : 1 1,85 : 0,925 : 0,925 : 0,925 NaOH is the limited reactant 10.2 NaOH : Na2CO3 2 : 1 1,85 : 0,925 0,925 mol of Na2CO3 is formed 10.3 We have 1 mol of CO2 and use 0,925 mol, thus 1 – 0,925 = 0,075 mol of CO2 is left

11.1 NH4NO3 : n =

𝑛 𝑁 = 30 80 = 0,375 mol

Na3PO4 : n =

𝑛 𝑁 = 50 164 = 0,305 mol

NH4NO3 : Na3PO4 3 : 1 0,375 : 0,125 thus NH4NO3 is limiting 11.2 NH4NO3 : (NH4)3PO4 : NaNO3 3 : 1 : 3 0,375 : 0,125 : 0,375 11.3 (0,305 mol of Na3PO4 given) – (0,125 mol of Na3PO4 used) = 0,18 mol Na3PO4 is left

12.1 CaCO3 : n =

𝑛 𝑁 = 100 100 = 1 mol

FePO4 : n =

𝑛 𝑁 = 45 151 = 0,298 mol

CaCO3 : FePO4 3 : 2 1 : 0,666 If all the CaCO3 is used, there is not enough FePO4

• present. Therefore the limited reactant is FePO4.

12.2 FePO4 : Ca3(PO4)2 : Fe2(CO3)3 2 : 1 : 1 0,298 : 0,149 : 0,149 thus 0,149 mol of Ca3(PO4)2 = 46,19 g is formed thus 0,149 mol of Fe2(CO3)3 = 43,51 g is formed

12.3 CaCO3 : FePO4 3 : 2 0,447 : 0,298 (1 mol of CaCO3 given ) – (0,447 mol of CaCO3 used) = 0,553 mol of CaCO3 over

9. 20 g of MgCℓ2 and 20 g of NaCℓ are mixed and dissolved in water to make a solution of 2 dm3. What is the concentration of the chloride ions in the solution?

n =

𝑛 𝑁

=

20 95 = 0,21 mol MgCℓ2

∴ n = 0,42 mol Cℓ- ions

c =

𝑜 𝑊

=

0,76 2

= 0,38 mol.dm-3

n =

𝑛 𝑁

=

20 58,5 = 0,34 mol NaCℓ

∴ n = 0,34 mol Cℓ- ions

• 10. Ammoniumnitraat is die hoofbestanddeel van “ANFO” (Ammonium nitrate fuel oils)

wat as plofstof in die mynindustrie gebruik word. Tydens die ontploffing vind die volgende reaksie plaas: 2NH4NO3 → 2N2(g) + 4 H2O(g) + O2(g) 800 g van die ammoniumnitraat word gebruik as plofstof. 10.1 Bereken die volume van die gas wat tydens die reaksie vrygestel word by STD.

n = 𝑛

𝑁

=

800 80 = 10 mol

2 mol ammoniumnitraat lewer (2 + 4 + 1) = 7 mol gas 10 mol ammoniumnitraat lewer 35 mol gas

n =

𝑊 𝑊

𝑛

35 =

𝑊 22,4

→ V = 784 dm3

• 10. Ammoniumnitraat is die hoofbestanddeel van “ANFO” (Ammonium nitrate fuel oils)

wat as plofstof in die mynindustrie gebruik word. Tydens die ontploffing vind die volgende reaksie plaas: 2NH4NO3 → 2N2(g) + 4 H2O(g) + O2(g) 800 g van die ammoniumnitraat word gebruik as plofstof. 10.2 Een van die gasse wat as ’n produk vrygestel word, is suurstof. Die suurstof onderhou verdere verbranding en maak dit moeilik om die ontploffing en vlamme te beheer. Bereken die massa suurstofgas wat tydens hierdie reaksie vrygestel word.

2 mol ammoniumnitraat lewer 1 mol suurstof 10 mol ammoniumnitraat lewer 5 mol suurstof

n =

𝑛 𝑁

5 = 𝑛

32

m = 160 g

• 11. Oktaan is een van die bestanddele van petrol. Tydens die verbranding van oktaan in die

enjin van ‘n motor, vind die volgende reaksie plaas: 2C8H18 + 25O2 → 16CO2(g) + 18H2O(g) 11.1 Watter massa koolstofdioksied sal vorm indien 80 g suurstof toegelaat word

• m met 23 g oktaan te reageer?

25 mol suurstof reageer met 2 mol oktaan 2,5 mol suurstof reageer met 0,2 mol oktaan

nsuurstof =

𝑛 𝑁 = 80 32 = 2,5 mol

noktaan = 𝑛

𝑁 = 23 114 = 0,202 mol

2 mol oktaan reageer om 16 mol koolstofdioksied te vorm 0,2 mol oktaan reageer om 1,6 mol koolstofdioksied te vorm

nkoolstofdioksied = 𝑛

𝑁

1,6 mol =

𝑛 44

m = 70,4 g

• 11. Oktaan is een van die bestanddele van petrol. Tydens die verbranding van oktaan in die

enjin van ‘n motor vind die volgende reaksie plaas: 2C8H18 + 25O2 → 16CO2(g) + 18H2O(g) 11.2 Watter een van die reaktante is die beperkende reaktant?

Suurstof

• 11. Oktaan is een van die bestanddele van petrol. Tydens die verbranding van oktaan in die

enjin van ‘n motor vind die volgende reaksie plaas: 2C8H18 + 25O2 → 16CO2(g) + 18H2O(g) 11.3 Watter volume gas word gevorm tydens hierdie reaksie indien 80 g suurstof toegelaat word om met 23 g oktaan te reageer?

2 mol oktaan reageer om 34 mol gas te vorm 0,2 mol oktaan reageer om 3,4 mol gas te vorm.

n =

𝑊 𝑊

𝑛

3,4 = 𝑊

22,4

V = 76,16 dm3

7. Die termiese reaksie behels die bereiding van ystermetaal uit 'n mengsel van aluminiumpoeier en yster(III)oksied. 'n Mengsel van 50 g yster(III)oksied en 50 g aluminium word gebruik. Die chemiese vergelyking vir die reaksie is soos volg: Fe₂O3 + 2Aℓ → 3Fe(s) + Aℓ2O3(s) 7.1 Bepaal die beperkende reagens.

7. Die termiese reaksie behels die bereiding van ystermetaal uit 'n mengsel van aluminiumpoeier en yster(III)oksied. 'n Mengsel van 50 g yster(III)oksied en 50 g aluminium word gebruik. Die chemiese vergelyking vir die reaksie is soos volg: Fe₂O3 + 2Aℓ → 3Fe(s) + Aℓ2O3(s) 7.2 Watter massa yster sal geproduseer word?

Power Gym

• p. 194,
• nos. 4, 5, 10

4. Magnesium burns in air to form magnesium oxide according to the following balanced equation: 2 Mg(s) + O2(g) → 2 MgO(s) If the percentage yield of this reaction is only 80%, calculate the mass of magnesium that has to be burnt to form 30 g of magnesium oxide.

40 (Mg) 24 0,9375 mol 0,94 22,5

• 10. Eggshells mostly consist of calcium carbonate. A sample of 5,5 g of eggshells reacts with

hydrochloric acid. The chemical equation for the reaction is: CaCO3 + 2 HCℓ → CaCℓ2 + CO2 + H2O 10.1 The reaction is completed and 1,12 dm3 of carbon dioxide is formed at STP. Calculate the mass of carbon dioxide formed during this reaction.

• 10. Eggshells mostly consist of calcium carbonate. A sample of 5,5 g of eggshells reacts with

hydrochloric acid. The chemical equation for the reaction is: CaCO3 + 2 HCℓ → CaCℓ2 + CO2 + H2O 10.2 Calculate the percentage purity of the eggshells.

Power Gym

• p. 200,
• nos. 2, 3, 7

• 2. Ammonium nitrite decomposes during heating to produce nitrogen gas and water vapour:

NH4NO2(s) → N2(g) + 2H2O(g) Calculate the total volume of gases produced at 819 K and 1 atm pressure when 128 g of ammonium nitrite undergoes this decomposition reaction.

3. When ammonium nitrate is heated, it undergoes a thermal decomposition reaction represented by the following equation. 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g) If 500 g of ammonium nitrate is heated: 3.1 Calculate the total volume of the gas produced at STP.

3. When ammonium nitrate is heated, it undergoes a thermal decomposition reaction represented by the following equation. 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g) If 500 g of ammonium nitrate is heated: 3.2 Calculate the total volume of the gas produced at 400 K and 150 kPa.

7. Potassium chlorate decomposes when heated to form potassium chloride and

• xygen, according to the following chemical

reaction: 2KCℓO3 → 2KCℓ + 3O2 A sample of 10 g potassium chlorate is heated and the oxygen gas formed, occupies a volume

• f 3,4 dm3 at room temperature (25 oC) at

1 atm pressure. Calculate the percentage yield for this reaction.